kakarotyjn
- 95
- 0
Let c be a singular k-cube and p:[0,1]^k \to [0,1]^k a 1-1 function such that p([0,1]^k ) = [0,1]^k<br />
<br /> and \operatorname{det} p'(x) \ge 0 for x \in [0,1]^k.If \omega is a k-form,show that <br />
<br />
\int\limits_c \omega = \int\limits_{c \circ p} \omega
Note that
\int\limits_c \omega = \int\limits_{[0,1]^k } {c*\omega } = \int\limits_{[0,1]^k } {(f \circ c)(\det c')dx^1<br /> <br /> \wedge ... \wedge dx^k }
\int\limits_{c \circ p} \omega = \int\limits_{[0,1]^k } {(c \circ p)*\omega } = \int\limits_{[0,1]^k } {(f \circ c<br /> <br /> \circ p)(\det (c \circ p)')dx^1 \wedge ... \wedge dx^k } = \int\limits_{[0,1]^k } {(f \circ c \circ p)((\det c') \cdot<br /> <br /> (\det p'))dx^1 \wedge ... \wedge dx^k }
did I deduce it right?If it's right,how to prove
Note that
\int\limits_c \omega = \int\limits_{[0,1]^k } {c*\omega } = \int\limits_{[0,1]^k } {(f \circ c)(\det c')dx^1<br /> <br /> \wedge ... \wedge dx^k }
\int\limits_{c \circ p} \omega = \int\limits_{[0,1]^k } {(c \circ p)*\omega } = \int\limits_{[0,1]^k } {(f \circ c<br /> <br /> \circ p)(\det (c \circ p)')dx^1 \wedge ... \wedge dx^k } = \int\limits_{[0,1]^k } {(f \circ c \circ p)((\det c') \cdot<br /> <br /> (\det p'))dx^1 \wedge ... \wedge dx^k }
did I deduce it right?If it's right,how to prove
Last edited by a moderator: